Base | Representation |
---|---|
bin | 10000110010101111111… |
… | …101011011101000110001 |
3 | 11002022200110220022001002 |
4 | 100302233331123220301 |
5 | 122401344324231423 |
6 | 2242050344451345 |
7 | 146242164434543 |
oct | 20625775335061 |
9 | 4068613808032 |
10 | 1154003352113 |
11 | 405456897807 |
12 | 1677a1479b55 |
13 | 84a8c0713b1 |
14 | 3dbd5781693 |
15 | 20041a2e228 |
hex | 10caff5ba31 |
1154003352113 has 2 divisors, whose sum is σ = 1154003352114. Its totient is φ = 1154003352112.
The previous prime is 1154003352107. The next prime is 1154003352133. The reversal of 1154003352113 is 3112533004511.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 1000384036864 + 153619315249 = 1000192^2 + 391943^2 .
It is a cyclic number.
It is not a de Polignac number, because 1154003352113 - 220 = 1154002303537 is a prime.
It is not a weakly prime, because it can be changed into another prime (1154003352133) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (23) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 577001676056 + 577001676057.
It is an arithmetic number, because the mean of its divisors is an integer number (577001676057).
Almost surely, 21154003352113 is an apocalyptic number.
It is an amenable number.
1154003352113 is a deficient number, since it is larger than the sum of its proper divisors (1).
1154003352113 is an equidigital number, since it uses as much as digits as its factorization.
1154003352113 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 5400, while the sum is 29.
Adding to 1154003352113 its reverse (3112533004511), we get a palindrome (4266536356624).
The spelling of 1154003352113 in words is "one trillion, one hundred fifty-four billion, three million, three hundred fifty-two thousand, one hundred thirteen".
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